Originally posted by JoeMcPhee
No, because ethanol has a density of 0.789 while a sugar solution has a density greater than 1.000, so when you measure the density of the solution, youíre measuring the average of both at the same time.

Maybe we just arenít understanding each other fully. I am working under the assumption that the sugar and ethanol would be concentrated at the same rate. Do you agree with this? If the sugar and ethanol maintain the same ratio then my numbers are spot on. If however the ice tends to hold onto ethanol and sugar at different rates, then my numbers for ABV could be wrong (either up or down depending on which ice holds onto more).

How can we resolve this central question? Math!

My calculations are based on the assumption that the ice melt has 13% of the strength of the original beer (both alcohol and sugar) because the OG is 13% that of the base beer (I use 6.5% in the calculations because the volume is only half of that of the base). This indicates that the ABV of the ice portion should be 10 (ABV of the base) x .13 = 1.3% ABV, and the real residual extract should be .036 (RE of the base) x .13 = .00468.

So we can think of the effective OG for the ice portion as .13 x 99 (OG for the base beer) = 1.013. What happens if we use an ABV calculator ( http://www.realbeer.com/spencer/attenuation.html ) to determine what the ABV of a beer with an OG of 1.013 and an FG of 1.003? Thatís right we get exactly what is predicted, 1.3% ABV. We also get a real extract of .0048 (just .00012 off the predicted).

All of this means that the ice holds the sugar and ethanol and almost exactly the same rate (if anything it holds onto the sugar slightly more), which in turn means that you can accurately estimate the strength of the concentrated beer on the ratio the FGs.

It is fine if you want to dispute this (although I’m not sure how you could), but please post something that says where you think my calculations are incorrect, and how you would correct them to show to "true" alcohol content.

Originally posted by JoeMcPhee
No, because ethanol has a density of 0.789 while a sugar solution has a density greater than 1.000, so when you measure the density of the solution, youíre measuring the average of both at the same time.

Maybe we just arenít understanding each other fully. I am working under the assumption that the sugar and ethanol would be concentrated at the same rate. Do you agree with this? If the sugar and ethanol maintain the same ratio then my numbers are spot on. If however the ice tends to hold onto ethanol and sugar at different rates, then my numbers for ABV could be wrong (either up or down depending on which ice holds onto more).

How can we resolve this central question? Math!

My calculations are based on the assumption that the ice melt has 13% of the strength of the original beer (both alcohol and sugar) because the OG is 13% that of the base beer (I use 6.5% in the calculations because the volume is only half of that of the base). This indicates that the ABV of the ice portion should be 10 (ABV of the base) x .13 = 1.3% ABV, and the real residual extract should be .036 (RE of the base) x .13 = .00468.

So we can think of the effective OG for the ice portion as .13 x 99 (OG for the base beer) = 1.013. What happens if we use an ABV calculator ( http://www.realbeer.com/spencer/attenuation.html ) to determine what the ABV of a beer with an OG of 1.013 and an FG of 1.003? Thatís right we get exactly what is predicted, 1.3% ABV. We also get a real extract of .0048 (just .00012 off the predicted).

All of this means that the ice holds the sugar and ethanol and almost exactly the same rate (if anything it holds onto the sugar slightly more), which in turn means that you can accurately estimate the strength of the concentrated beer on the ratio the FGs.

It is fine if you want to dispute this (although Iím not sure how you could), but please post something that says where you think my calculations are incorrect, and how you would correct them to show to "true" alcohol content.

That assumption is wrong... they will not freeze out at the same rate... math doesn’t get you past physics.

Originally posted by puzzl
Iím going to go ahead and ice some crappy beer and drink the frozen solution and let you guys know what happens.

I think youíre in for a treat!

Considering eisbier is a style using freezing to greater concentrate alcohol then one might assume that under certain or most conditions icing will produce disproportionate icing of water and alcohol.

I’ve decided on an 06 Black Chocolate Stout. I’m going to pour the beer into a bowl and stick it in the freezer along with a small strainer, and will skim the ice every 10-15 minutes until I’ve got about 50% off. That sound like it make sense to people who know more about this stuff than me?

Originally posted by puzzl
Iím going to go ahead and ice some crappy beer and drink the frozen solution and let you guys know what happens.

I think youíre in for a treat!

Considering eisbier is a style using freezing to greater concentrate alcohol then one might assume that under certain or most conditions icing will produce disproportionate icing of water and alcohol.

Iíve decided on an 06 Black Chocolate Stout. Iím going to pour the beer into a bowl and stick it in the freezer along with a small strainer, and will skim the ice every 10-15 minutes until Iíve got about 50% off. That sound like it make sense to people who know more about this stuff than me?

about half way through this thread a suspicious character shares his thoughts on this subject. the same person also enjoys iced mill st tankhouse ale because he doest not have arogant bastard readily availible.